# Moment of inertia around two axes

In the this chapter of an online pdf we are given an equation for the deflection of a beam: $$\frac{d^2y}{dx^2}=\frac{\overline{M}}{E I}$$ where $E$ is the modulus of elasticity, $\overline{M}$ is the moment as function of $x$ and $I$ is the moment of inertia. It works nicely for a beam with one free end as the beam would be rotating around a single axis. However when they get to the worked example with both ends fixed they leave the moment of inertia uncalulated (left just as $I$). I am wondering how one would calculate the moment of inertia for such a beam. Is it even defined?

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Note that the $I$ in $T=I\,\omega$ is completely different from the $I$ in $M = E\,I y''$. Apparently this Prof. in the department of Mathematics in NJIT does not have an engineering background. –  ja72 Nov 11 '13 at 14:41
A flexed beam is not rotating about a fixed axis. The end if the beam is not rotating and the further away you go the more rotation you have, in a non linear fashion. A rotation as defined in dynamics implies deflection simply proportional to distance. –  ja72 Nov 11 '13 at 15:09

The area moment of inertia for beams is defined from the geometry of the beam cross section and not from the end conditions. It is defined as $I_{xx} = \int (y^2+z^2)\,{\rm d}A$ and has units of LENGTH4. The $x$ and $y$ values are usually displacements from the neutral axis which lies along the centroid.

The same equation holds regardless of the end conditions. The end conditions define the constants of integration when getting $y(x)$.

## Example

Rectangular beam with elasticity $E$ and length $L$ is loaded in the middle by force $P$, and fixed at the ends.

1. Look up in the lists of area moments $I_{xx} = \frac{b^2+h^2}{12}$ where $b$ is base of rectangle cross section and $h$ the height.
2. The internal moment a distance $x$ from one of the ends (I call A) is $M(x)=M_A - x\,A_y$ with $A_y$ the vertical reaction force, and $M_A$ the reaction moment at A. Force and moment exist in order to enforce the no displacement and no slope at A constraints.
3. The deflection shape is $$y(x) =C_0 + C_1 x + \int \int \frac{M(x)}{E I}\,{\rm d}x\,{\rm d}x = C_0 + C_1 x +\frac{x^2 \left(3 M_A + x\,A_y \right)}{6 E I}$$
4. The end conditions at A are $y(0)=0$ and $y'(0)=0$ (with $y'$ denoting slope) yielding $C_0=0$ and $C_1=0$ $$y(x) = \frac{x^2 \left(3 M_A + x\,A_y \right)}{6 E I}$$
5. In the middle of the beam the slope must be zero (due to symmetry) and thus $y'(\frac{L}{2})=0$ yielding $$M_A = \frac{A_y L}{4}$$.
6. The total reaction force must be equal to the applied load and thus $A_y=\frac{P}{2}$
7. The deflection shape is now $$\boxed{ y(x) = \frac{P\,x^2 \left(3 L-4 x\right)}{48\;E\,I} }$$
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I am not an expert in this and since I couldn't comment I'll just leave what I have to say here.

Look up the Parallel Axis Theorem:

$I_A = I_{COM} + Mh^2$

Where $I_A$ is the moment of interia at any point, $A$, of the object, $M$ is the mass of the object and $h$ is the distance from the COM (Centre of Mass) to the point $A$. $I_{COM}$ is the moment of intertia around the centre of mass.

Ignore $d$ from the diagram; I just wanted to show you the typical set up of the problem.

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You are confusing mass moment of inertia, with area moment of inertia. The first is used in dynamics, and the second in statics. –  ja72 Nov 11 '13 at 14:29
Are you saying that because of the diagram? As in: it should not be a 3D object. I should have found a better example. Or maybe the Parallel Axis Theorem isn't the correct method for this question? –  PPG Nov 11 '13 at 14:36
Yes, but with area, not mass. $I_A = I_C + A\,d^2$ where $A$ is the cross section area and $I_C$ the area moment about the centroid. –  ja72 Nov 11 '13 at 14:39